Question

我想使用c和pic +光电二极管制作一个人我的代码使用c：假设我们在门前有两个二极管来计算来电人数和一个在门后面减少左边人数，我的问题由代码：

while(1)
{
//suppose we declare a count variable to hold the count 
    if(d1 == 1) //suppose it's the first diode and when it's cut it's value will be one
    {
        count++;
    }
    if(d2 == 1)// represent d2
    {
        count--;
    } // now my problem is when the count is increase by one it will decrease also by one
}

你能帮忙吗？

Answer 1

我并不完全确定你是如何进行此设置的，但我认为您需要在阅读后清除变量。因此，您的代码需要看起来更像这样：

int inOut = 0; // Positive for in negative for out

while(1)
{
    if(d1 == 1)
    {
        if(d2 == 1)
        {
            count += inOut;
            d1 = 0;
            d2 = 0;
        }
        else
        {
            inOut = 1;
        }
    }
    else if(d2 == 1)
    {
        inOut = -1;
    }
}

Answer 2

这应该有用，它不优雅，但它是：

int Entered = 0;
int Exited = 0;

while(1)
{  
  if(Exited == 0 && d1 == 1 && d2 == 0)
  {
    Entered = 1;
  }

  if(Entered == 1 && d1 == 0 && d2 == 1)
  {
    count++;
    Entered = 0;
  }

  if(Entered == 0 && d1 == 0 && d2 == 1)
  {
    Exited = 1;
  }

  if(Exited == 1 && d1 == 1 && d2 == 0)
  {
    count--;
    Exited = 0;
  }
}

如果你不明白，请问我。

Answer 3

亲爱的，你的回答是行不通的，因为有人进入它会削减第二个二极管，它会减少计数器，但最后我想我得到了答案：我们将改变每个二极管的位置，使其彼此相邻并延迟 //假设我们有3个变量，in，out和finally count

while(1)
{
if(D1 == 1) //somone come
{
count++;
D1 = D2 = 0;
}
delay(time)
if(D2 == 1)
{
count--;
D1 = D2 = 0;
}
}

人们使用Pic和c计数器

3 个答案: